How do you solve #4-2(2+4x)=x-3#?

First, expand the terms within parenthesis on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#4 - color(red)(2)(2 + 4x) = x - 3#

#4 - (color(red)(2) * 2) - (color(red)(2) * 4x) = x - 3#

#4 - 4 - 8x = x - 3#

#0 - 8x = x - 3#

#-8x = x - 3#

Next, subtract #color(red)(x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(red)(x) - 8x = -color(red)(x) + x - 3#

#-1color(red)(x) - 8x = 0 - 3#

#(-1 - 8)x = -3#

#-9x = -3#

Now, divide each side of the equation by #color(red)(-9)# to solve for #x# while keeping the equation balanced:

#(-9x)/color(red)(-9) = (-3)/color(red)(-9)#

#(color(red)(cancel(color(black)(-9)))x)/cancel(color(red)(-9)) = (-3 xx 1)/color(red)(-3 xx 3)#

#x = (color(red)(cancel(color(black)(-3))) xx 1)/color(red)(color(black)(cancel(color(red)(-3))) xx 3)#

#x = 1/3#

Firstly, distribute the bracket.

#rArrcancel(4)cancel(-4)-8x=x-3#

#"subtract x from both sides"#

#-8x-x=cancel(x)cancel(-x)-3#

#rArr-9x=-3#

#"divide both sides by - 9"#

#(cancel(-9) x)/cancel(-9)=(-3)/(-9)#

#rArrx=1/3#

#color(blue)"As a check"#

Substitute this value into the equation and if the left side equals the right side then it is the solution.

#4-2(2+4/3)=4-2(10/3)=4-20/3=-8/3#

#1/3-3=1/3-9/3=-8/3larrcolor(red)" right side"#

#rArrx=1/3" is the solution"#